Lesson Explainer: Two-Slit Interference Physics • 9th Grade

In this explainer, we will learn how to calculate the positions of points of maximum and minimum intensity in interference patterns generated by double slits.

To begin our experiment with light, we need two small parallel slits in some barrier. Next, we need coherent light waves, that is, waves of the same wavelength and a constant phase difference, to be traveling toward them. To see what happens to the light as it passes through the slits, a screen is placed some distance away from the other side of the barrier. This setup is shown in the figure below. It is important to understand how the light waves are represented here. The equidistant lines represent a bird’s eye view of the waves. The gray lines represent the peaks of the waves as they travel left to right across the page and toward the double slit.

What you would expect to see are two bright spots directly opposite the slits, as the barrier is preventing the light illuminating the rest of the screen. However, this is not what is seen. Instead, there is a series of bright spots, not just two, along the whole screen. So, what is happening here? What is happening here is, in one word, interference.

First, consider just one of the slits. The waves passing through it will spread out, or diffract, as light passes through the aperture. Light will diffract whenever it passes an obstacle, such as a corner, or, in this case, a slit. Remember that diffraction is actually occurring at both slits, so as the waves spread out from both, they overlap and interfere with each other. This is what creates the pattern of bright and dark spots seen on the screen.

To understand why this interference creates this pattern, we need to understand the principle of superposition. This is a rather complicated name for a relatively straightforward idea. To explain this principle, we need to move away from our bird’s eye view of the light waves shown in previous figures and represent the wave graphically as a curve again.

Definition: Principle of Superposition

Looking at the figure above, if we have two waves overlapping, with their peaks and troughs lined up, they are said to be in phase. The principle of superposition tells us that the waves will constructively interfere with each other. The resulting wave can be thought of as the addition of the two peaks and the addition of the two troughs, leading to a wave of larger amplitude.

However, if the two waves are out of phase, then the peaks of one wave line up with the troughs of the other and, by the principle of superposition, they will destructively interfere. This can be thought of as the addition of the troughs and the peaks. Since the troughs are below the horizontal, the displacement is negative. Because we are adding a negative, we are in fact just subtracting the troughs from the peaks. This means the two waves will cancel each other out.

Returning to our two-slit experiment, the principle of superposition tells us that the bright spots (or points of maximum intensity) on the screen correspond to places where the overlapping light is constructively interfering and that the dark spots (or points of minimum intensity) correspond to places where overlapping light is destructively interfering.

The next step is to quantify this interference and get an equation that describes what is happening. As with many problems in physics, a good place to start is a labeled diagram. Looking at the figure below, we are considering a specific point on the screen and the two paths the light takes from the slits to that point.

First, we introduce the concept of path length difference.

Definition: Path Length Difference

If we take π‘‘οŠ§ to be the distance traveled by the light coming out of the first slit and π‘‘οŠ¨ to be the distance traveled by the light coming out of the second slit, then the path length difference is just the absolute value of the difference between these two numbers.

Or rather, it is how much further the light from one slit has to travel compared to light from the other to reach the same point on the screen. It is given by the equation |π‘‘βˆ’π‘‘|=Δ𝑑.

For a dark spot, it is this difference that means the waves arrive out of phase and destructively interfere. For a bright spot, the waves arrive in phase and constructively interfere.

We can define 𝑑 as the distance between the centers of the slits. We can then construct a right triangle using 𝑑, πœƒ, and Δ𝑑, as seen in the figure. Remember that Δ𝑑 is the path length difference of the two lines and so corresponds to the base of the triangle. This is an important step, as we can now use trigonometry of right triangles to get an equation.

With 𝑑 as the hypotenuse, we know that π‘‘πœƒ=Δ𝑑.sin

At this point we need to outline one big assumption we have made, which is that the two paths taken by the light are parallel. We need to make this assumption so that we can construct the right triangle and form our equation. The two lines are clearly not parallel, as they meet at the point on the screen where the interference is occurring. This assumption requires that πœƒ=πœƒ=πœƒοŠ§οŠ¨. It is therefore best to understand πœƒ as the angle at which the paths of both waves are tilted upward. This approximation is good enough, as the distance to the screen is large, specifically much larger than distance 𝑑.

Looking at our equation, we can see that 𝑑 is a fixed value and will not change. This means the path length difference changes as the direction of the light, measured by πœƒ, also changes. It is this change that causes the pattern of dark and light spots on the screen.

We now need to add to our equation a way to quantify phase difference, so we have an equation that describes constructive interference (when the waves are in phase) and another equation that describes destructive interference (when the waves are out of phase).

The phase difference between the two waves is measured in wavelengths. It is important to know that both waves will have the same wavelength, πœ†. At a phase difference of 0πœ†, clearly the two waves will be in phase and will constructively interfere as their peaks/troughs line up. However, since the waves are periodic and repeat every πœ†, the waves will line up again at 1πœ†. In fact, any phase difference described by a multiple of π‘›πœ† (where 𝑛 is any integer) will result in constructive interference and a bright spot on the screen.

Similarly, we said when the peaks of one wave line up with the troughs of the other, they will destructively interfere and there will be a dark spot on the screen. To line the waves up this way, we need a phase difference of 12πœ†. But if we again consider the periodic nature of the waves every πœ†, there will then be destructive interference at 32πœ†, 52πœ†, and so on. In other words, for any whole number 𝑛, destructive interference will occur at any phase difference following the rule 𝑛+12οˆπœ†.

We are finally in a position to get an equation for both constructive and destructive interference. Key to this is understanding that the phase difference and the path length difference are equal to each other. This is true because if the waves arrive at different times, then the peaks/troughs will then line up differently and may or may not have a phase difference. So, constructive interference has the equation π‘‘πœƒ=π‘›πœ†,sin where 𝑛=0 represents the center bright spot, the first point of maximum intensity. Destructive interference has the equation π‘‘πœƒ=𝑛+12οˆπœ†,sin where 𝑛=0 represents the first dark spot above the center, the first point of minimum intensity. Now, we work through some examples that rely on the equations for constructive and destructive interference. Below is a helpful diagram detailing what the different values of 𝑛 correspond to on the screen. A larger 𝑛 means you are further from the center of the screen as the path length difference must be larger.

Next, we work through some solved examples. Often, questions will talk about fringes rather than about bright/dark spots. The ideas and equations are all the same; it is just a better representation of what is seen on the screen in a real-life experiment. The questions will often talk about the center of the fringes. It is this distinction that allows us to treat these fringes as spots, as the centers of the fringes correspond to where the spots would be in our simplified diagrams.

Example 1: Calculating the Wavelength of Light for Constructive Interference

Light passes through a sheet in which there are two parallel narrow slits 12.8 μm apart. The light from the slits is incident on a screen parallel to the sheet, where a pattern of bright and dark fringes is observed. A line 𝐿 runs perpendicular to the surface of the sheet and the direction of the slits. The line 𝐿 intersects the central bright fringe of the pattern on the screen. The angle between 𝐿 and a line that intersects the center of the bright fringe closest to the central bright fringe is 3.09∘. What is the wavelength of the light? Give your answer to the nearest nanometre.

Answer

A good first step with a question like this is to assess what we know and what we need to find out. We are dealing with a bright fringe, so we need the equation describing constructive interference: π‘‘πœƒ=π‘›πœ†.sin

The first number given to us is the distance between the slits, so we know that 𝑑=12.8.ΞΌm

The next important part is the angle and what it is representing. The question tells us that it is the angle up to the fringe closest to the central bright fringe (i.e., up from the middle of the screen). This tells us we are at the first order or rather that 𝑛=1 and also gives us that πœƒ=3.09.∘

The next step is to rearrange the equation so that πœ† is the subject. Simple algebra tells us that πœ†=π‘‘πœƒπ‘›.sin

Finally, all that is left to do is substitute in the values, taking care to convert all units to metres. The result is πœ†=12.8Γ—10Γ—(3.09)1.sin

Our final answer is πœ†=6.9Γ—10=690mnm when we convert it back to nanometres.

Next is a similar, slightly longer example.

Example 2: Calculating the Angles between Fringes in Two-Slit Interference

Light with a wavelength of 597 nm passes through a sheet in which there are two parallel narrow slits 7.64 μm apart. The light from the slits is incident on a screen parallel to the sheet, where a pattern of light and dark fringes is observed. A line 𝐿 runs perpendicular to the surface of the sheet and the direction of the slits, intersecting the central bright fringe of the pattern on the screen. Two lines, line I and line II, intersect 𝐿 at the position of the sheet. Line I intersects the center of the dark fringe closest to the central bright fringe and line II intersects the center of the bright fringe closest to the central bright fringe. Both lines I, and II are on the same side with respect to line 𝐿. What is the angle between line I and line II? Give your answer to one decimal place.

Answer

We are dealing with both a bright fringe and a dark fringe, so we need two equations describing constructive and destructive interference. These are π‘‘πœƒ=π‘›πœ†,sin for constructive interference, and π‘‘πœƒ=𝑛+12οˆπœ†,sin for destructive interference.

It will be helpful to draw a diagram to find out what we know and what we need to find out.

So, to find πœƒdiff, we need to find the difference between πœƒοŒΊ and πœƒοŒ».

From the question, we have that 𝑑=7.64,πœ†=597.ΞΌmnm

So, the only unknown left in our equations, other than the angles, is 𝑛. Looking at the diagram, we are at the closest dark fringe (line I) to the central bright fringe and the closest bright fringe (line II) to the central bright fringe. This means we have our bright fringe at 𝑛=1 and the first dark fringe at 𝑛=0.πœƒοŒΊ is the angle between line I and 𝐿, and so its location is defined by the equation for destructive interference, where we have to add 12 to 𝑛. Firstly, we should rearrange the destructive interference equation, remembering to use inverse sine: πœƒ=𝑛+ο‡πœ†π‘‘οŽ.sin

When substituting into the equation, being careful to convert to metres, we get πœƒ=ο€Ύ(0.5)Γ—597Γ—107.64Γ—10.sin

This gives the result πœƒ=2.2.∘

Next is πœƒοŒ», which is the angle between line II and 𝐿, and so its location is defined by the equation for constructive interference. Rearranged for πœƒοŒ», it is πœƒ=ο€½π‘›πœ†π‘‘ο‰.sin

When substituting into the equation, being careful to convert to metres, we get πœƒ=ο€Ύ597Γ—107.64Γ—10.sin

This gives the result πœƒ=4.4.∘

Finally, since we have established that πœƒ=πœƒβˆ’πœƒ,diο¬€οŒ»οŒΊ our final answer is πœƒ=4.4βˆ’2.2πœƒ=2.2.diffdiο¬€βˆ˜βˆ˜βˆ˜

This next example requires more critical thinking in order to work out exactly what the question is telling us.

Example 3: Calculating the Number of Fringes in Two-Slit Interference

Light with a wavelength of 563 nm passes through a sheet in which there are two parallel narrow slits 8.38 μm apart. The light from the slits is incident on a screen parallel to the sheet, where a pattern of light and dark fringes is observed. A line 𝐿 runs perpendicular to the surface of the sheet and the direction of the slits. The line 𝐿 intersects the central bright fringe of the pattern on the screen. How many bright fringes will be present on a screen that can extend, without limit, either side of 𝐿?

Answer

This is a difficult question conceptually, but it uses a simple equation. Let’s start with writing out what we know and, from that, what equation we need to work with. We are given πœ†=563,𝑑=8.38.nmΞΌm

To work out what to do next, we need to carefully consider what the question is telling us. It is asking us to find the total number of fringes on the screen without limit. This tells us two things.

First, it tells us that the thing we are trying to calculate is 𝑛, specifically 2×𝑛. Remember that 𝑛 is essentially what number fringe you are at, counting up from the center. So, to get the total fringes on both sides, we need to double this number.

Secondly, the β€œwithout limit” part tells us that we are dealing with an angle πœƒ. Specifically, we need to have πœƒ=90.∘

This is because we need to consider the furthest possible fringe on the screen, which will be at infinity. As we move further and further away from the central fringe, πœƒ will get larger and larger. The path line of the light, drawn from the slits to the screen, will tilt more and more toward the vertical position as πœƒ and 𝑛 increase. So, at infinity, we will have reached the final value of 𝑛, the path line of the light will be vertical, πœƒ will be 90∘, and we will have checked all possible fringes on the screen. The word infinity might be a bit off-putting here, but just remember that we want to make sure we are not missing any fringes. So, we need to move our path line as far across the screen as it will go.

With all that established, we now know what equation to use: sinπœƒ=π‘›πœ†π‘‘.

As we are dealing with bright spots, there must be constructive interference. Rearranged for 𝑛, using basic algebra, the equation becomes 𝑛=π‘‘πœƒπœ†.sin

Substituting in, being careful to convert into metres, we get 𝑛=8.38Γ—10Γ—(90)563Γ—10.sin

This gives a value of 𝑛=14.88454….

As discussed before, our answer is actually 2𝑛 because we have only counted the half of the fringes that is above the central fringe and not the other half below: 2𝑛=29.76909….

This, however, is not the final answer. Your instinct may be to round this number up to 30 or even 29.8. This would be wrong. To start, you cannot have 0.8 of a fringe, so a whole number is required for a correct answer. Secondly, we cannot round up because that would be adding additional fringes to our result. So, we must round our result down. If we reword the question to ask how many whole fringes there are, then it becomes clearer why we must round down. So, the final answer is fringes=29.

What if instead we want to find the location of the bright spots based on the distance from the center on the screen and not the angle πœƒ from the center point? Just like before, we will create a labeled diagram and then rely on trigonometry to construct an equation.

Looking at the figure, we have π‘¦οŠ, the vertical distance to the 𝑛th bright spot on the screen. This forms the height of the right triangle. There is also 𝐿, the distance from the slits to the center of the screen. This forms the base of the right triangle. It is also useful to define 𝐻 as the hypotenuse of the triangle and πœƒοŠ as the 𝑛th angle up from the center.

At this point, we have made an assumption. Notice, on the diagram, the slits appear much smaller than before. This is because we have zoomed out from the previous perspective in order to demonstrate that slit distance 𝑑 at this scale is much smaller than 𝐿. This means we approximate the light as coming from the same point on the barrier where the line 𝐿 starts. This simplifies the problem and allows us to construct the right triangle. This assumption is okay because in experiments involving slit interference, 𝑑 is of a similar magnitude to the wavelength of light used. This provides maximum diffraction and hence a clearer interference pattern on the screen. Since the light used is visible light, distance 𝑑 is of the scale of hundreds of nanometres and our assumption is therefore a good enough approximation.

Using the trigonometry of the right triangle, we can get two equations, one involving 𝐻 and the other involving 𝐿: tansinπœƒ=𝑦𝐿,πœƒ=𝑦𝐻.

Next, we rely on a second assumption, specifically the small angle approximation. We have established that the screen has been placed some large distance, 𝐿, away from the barrier in the initial setup. This means angle πœƒοŠ is very small, and 𝐿 is very close in value to 𝐻. We can therefore approximate that tansinπœƒ=πœƒοŠοŠ and hence can define our equation as sinπœƒ=𝑦𝐿.

The final step is to remove the πœƒ part of the equation, as we want our bright spot location to be defined by distance only. This is why we changed out the tan function for a sine, as we can now take the already known equation for constructive interference, sinπœƒ=π‘›πœ†π‘‘, and set it equal to the equation, yielding 𝑦𝐿=π‘›πœ†π‘‘.

This can be easily rearranged to make π‘¦οŠ the subject: 𝑦=π‘›πœ†πΏπ‘‘.

This gives us the equation we were looking for. It describes the locations of the bright spots (locations of constructive interference) in terms of distance from the center.

Below are some worked examples for the equations we have derived and the concepts we have introduced.

Example 4: Calculating the Distance between Fringes in Two Slit Interference

Light with a wavelength of 604 nm passes through a sheet in which there are two parallel narrow slits 9.44 μm apart. The light from the slits is incident on a screen parallel to the sheets, 1.25 m away, where a pattern of light and dark fringes is observed. A line 𝐿 runs perpendicular to the surface of the sheet and the direction of the slits. The line 𝐿 intersects the central bright fringe of the pattern on the screen. What is the distance on the screen from 𝐿 to the center of the bright fringe nearest to the central bright fringe? Give your answer to the nearest centimetre.

Answer

The best way to start a question like this is to write down the relevant equation and the values the question has given us. We have been given that πœ†=604,𝑑=9.44,𝐿=1.25.nmΞΌmm

Since the question wants us to find the distance on the screen, we need the equation 𝑦=π‘›πœ†πΏπ‘‘, where π‘¦οŠ is the distance on the screen from the central bright spot to the 𝑛th bright spot. We also have been told the value of 𝑛, as we are at the nearest bright fringe to the central fringe. So, 𝑛=1.

The next step is to simply substitute in the values, taking care to convert all units to metres, to get the answer: 𝑦=1Γ—604Γ—10Γ—1.259.44Γ—10=0.08.m

Finally, we must convert to centimetres as required by the question: 𝑦=8.cm

This next example is mathematically simple but requires critical thinking to get the right result.

Example 5: Comparing the Wavelengths of Different Interference Patterns

Light of two different wavelengths passes through a sheet in which there are two parallel narrow slits. The light from the slits is incident on a screen parallel to the sheet, where a pattern of light and dark fringes is observed. A line 𝐿 runs perpendicular to the surface of the sheet and the direction of the slits. The line 𝐿 intersects the central bright fringe of the pattern on the screen. The distance on the screen from 𝐿 to the center of the bright fringe of the shorter wavelength nearest to the central bright fringe is 5.55 cm. The distance on the screen from 𝐿 to the center of the bright fringe of the longer wavelength, nearest to the central bright fringe is 7.25 cm. What is the ratio of the longer-wavelength light to that of the shorter-wavelength light? Give your answer to two decimal places.

Answer

This question only gives us two values. They represent the distance between the first fringe and the central fringe for two separate patterns and hence two separate locations for the first bright fringes. Specifically, 𝑦=𝑦=5.5,shortscm for the shorter wavelength, and 𝑦=𝑦=7.25,longerlcm for the longer wavelength. At this point, we can recognize that the equation to use is 𝑦=π‘›πœ†πΏπ‘‘, as π‘¦οŠ is the distance to the 𝑛th fringe up from the center. This means we have two equations, one for each wavelength: 𝑦=π‘›πœ†πΏπ‘‘,𝑦=π‘›πœ†πΏπ‘‘.ssll

Here, we need to remember that these equations are describing the distance to the first bright fringes for both patterns. This means that 𝑛=1: 𝑦=1Γ—πœ†πΏπ‘‘,𝑦=1Γ—πœ†πΏπ‘‘.ssll

Remembering that the wavelengths are what we need, we can rearrange to make them the subjects: πœ†=𝑦𝑑𝐿,πœ†=𝑦𝑑𝐿.ssll

The next problem we run into is that we do not know any other parts of the two equations. We can get around this because the question is for the ratio of the two wavelengths. Keeping in mind that 𝑑 and 𝐿 are the same for both equations, we can combine the two to form one equation: πœ†πœ†=.lsls

Now, the values we do not know are canceled out by this division and we are left with πœ†πœ†=𝑦𝑦.lsls

The final answer is πœ†πœ†=7.255.55=1.31.ls

Key Points

  • Coherent light waves passing through a double slit will diffract (and hence interfere with each other), creating a pattern of bright/dark spots across a screen.
  • For the 𝑛th bright spot (the maxima), constructive interference occurs at angles defined by the equation π‘‘πœƒ=π‘›πœ†sin.
  • For the 𝑛th dark spot (the minima), destructive interference occurs at angles defined by the equation π‘‘πœƒ=𝑛+12οˆπœ†sin.
  • The equation 𝑦=π‘›πœ†πΏπ‘‘οŠ defines the location of the 𝑛th bright spot based on the distance from the screen center.

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